# Working out frictional force

My friend and I did an experiment, and we are to work out the frictional force, but have gone through two different routes with two different answers. I feel like solution "1" is correct, but can't quite figure out why solution "2" isn't correct. Maybe it is! Could you comment please?

                 m2⇓                     m1⇓


The mass $m_1$ falls $(h)$ to the ground and displaces the block $m_2$ to the right by $(h)$. This takes time $t$. The final velocity is calculated using $v=2\frac st$. To calculate the Force of Friction between $m_2$ and the table, we came up with

Solution 1

Finding the gravitational potential energy of $m_1$ via $m_1gh$, then deduct this from the kinetic energy of $m_2$, which was calculated via $0.5m_2v^2$. This gives the work done against friction, and when work done is known, $\frac {work}{distance}=$ Force of Friction

Another way of getting to this figure was finding the weight of $m_1$, thus knowing what force was acting on it, then calculating it's acceleration as it fell via $\frac{2s}{t^2}$. Once the acceleration was known, this was multiplied by $m_1$ to get the Net Downwards Force. This was subtracted from the Weight to find the opposing force, the Force of Friction.

The answers weren't identical, but close.

Solution 2

We know how far $m_2$ displaces, and over what time, so we know it's acceleration. $m_2a$ = Net Force. The difference between the Net Force and applied force (weight of $m_1$) gives the Frictional Force.

I feel like the error in Solution 2 stems from the calculation of the Net Force, by using $m_2$, but can't articulate it well to convince my friend.

What do you think?

• What is GPE? Please avoid using abbreviations. Mar 4, 2017 at 16:04
• Okay, thanks. I removed the abbreviations and fixed some of the equations that were changed in the reformatting! Mar 4, 2017 at 16:45
• How do you know the acceleration of m2 just from knowing displacement and time? Acceleration is change in velocity over time, so you would need to know its final velocity. Likewise, in the first solution, how do you know v, when calculating kinetic energy? Mar 4, 2017 at 17:00
• Benitok, I believe from the equation of motion (where acceleration is constant, s is displacement, t is time. and initial velocity u=0), then $s=0.5(u+v)t$ we can rearrange to calculate final velocity as $v=\frac{2s}t$? Mar 4, 2017 at 17:48

Both methods should give the same correct answer, if applied correctly. One uses work done = force x distance, the other uses net force = mass x acceleration.

SOLUTION 1 is correct in theory. However, you have missed out the kinetic energy of mass $m_1$. The PE lost by $m_1$ is equal the the work done against friction plus the KE gained by both masses.

If you use this method you must measure the final velocity of $m_1$ or $m_2$ (they have the same velocity) when $m_1$ has fallen through height $h$.

SOLUTION 2 seems to be the same method as "another way" mentioned under Solution 1, except applied to $m_2$ instead of $m_1$. The distances moved by $m_2$ and $m_1$ are the same, as are their accelerations, because they are attached by an inextensible string.

This method is also correct in theory : the net force $m_2$ or $m_1$ equals its mass times its acceleration. However, the applied force is the tension in the string, which is not the weight of $m_1$, it is $m_1(g-a)$ where $a$ is the common acceleration of both masses. If $m_1$ was in free fall $(a=g)$ then the tension in the string would be zero.

To use this method you must measure acceleration $a$, eg from the time which it takes $m_1$ to fall from rest through height $h$ - or for $m_2$ to move through a distance $h$ from rest. The total force on the 2 masses is $m_1g-F$ where $F$ is the friction force. The total mass being accelerated is $m_1+m_2$. Therefore
$m_1g-F=(m_1+m_2)a$
from which $F$ can be calculated, after $a$ has been measured.

• Thanks for that, some interesting points to consider, everything seems a bit clearer given that direction. I'm going to have a think about what would happen if there was no friction between $m_2$ and the table, especially with regards to the tension. I feel like there would have to be some tension in order for $m_2$ to move, yet $a$ for both objects should $= g$? Mar 6, 2017 at 10:51
• If the objects are connected by the string then $a \ne g$. Without friction $(m_1+m_2)a=m_1g$. If $m_1$ is in free fall then either the string has been cut ($a=g$ exactly, in which case $m_2$ does not accelerate at all) or $m_1 \gg m_2$ (then $a \approx g$, both particles accelerate, and the tension in the string is $m_2a$). Mar 6, 2017 at 15:06